About me

Outline

• Why be normal?

• GLM with binary data

• GLM with count data

What to expect during this workshop

These workshop will run for 2 hours each.

• Combines slides, live coding examples, and exercises for you to participate in.

• Ask questions in the chat throughout!

What to expect during this workshop

What you need for this workshop

• You are comfortable with simple operations in R.

• You know how to perform linear regression.

Introduce yourselves

https://blog.slido.com/virtual-icebreakers/

Workshop resources

Data

Guidotti, E., Ardia, D., (2020), “COVID-19 Data Hub”, Journal of Open Source Software 5(51):2376, doi: 10.21105/joss.02376.

Data

How to download:

covid <- readr::read_csv(
  "https://raw.githubusercontent.com/nrennie/f4sg-gams/main/data/covid.csv"
  )

• See data/ folder on GitHub for pre-processing.

Recapping general linear models

OLS

Limitations of general linear models

Q. What are the assumptions of a general linear model?

Model assumptions

The equation of a linear model (lm) is given by:

\[ y_i = \beta_0 + \beta_1 x_i + \epsilon_i \]

where:

The linear regression line is the most likely line given your data if we assume each data point comes from a hypothetical bell curve centered on the regression line

Normal distribution

Normal distribution of the residuals

When using our residuals to calculate standard error, confidence intervals and statistical signifcance these are assumed to be drawn from a normal distribution with mean zero and a constant variance.

The linear model equation

Another way to write the lm equation is:

\[ y_i \sim N(\mu = \beta_0 + \beta_1 X_i, \sigma^2) \] :::{.fragment} Which literally means that \(y_i\) is drawn from a normal distribution with parameters \(\mu\) (which depends on \(x_i\)) and \(\sigma\)(which has the same value for all measures of \(Y\)). :::

Real data

How often do these assumptions hold true?

good model Count data Challenger data?

Assumption testing

Residual plots?

car::vif(mod)
lmtest::bptest(mod)

sresid <- residuals(janka_gamma, type = "pearson")
bptest(janka_gamma, ~ sresid)


shapiro.test(residuals(model))

Theoretical foundations

Probability distributions and exponential families

The link function

Let’s start with linear models…

Janka timber hardness

This regression analysis uses data from the Australian forestry industry recording wood density and timber hardness from 36 different samples

• Timber hardness is quantified on the “Janka” scale

• The ‘amount of force required to embed a 0.444” steel ball into the wood to half of its diameter’.

Fitting the general linear model

janka_ls <- lm(hardness ~ dens, data = janka)

summary(janka_ls)
janka |> 
  ggplot(aes( x = dens, y = hardness))+
  geom_point()+
  geom_smooth(method = "lm")

Call:
lm(formula = hardness ~ dens, data = janka)

Residuals:
    Min      1Q  Median      3Q     Max 
-338.40  -96.98  -15.71   92.71  625.06 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) -1160.500    108.580  -10.69 2.07e-12 ***
dens           57.507      2.279   25.24  < 2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 183.1 on 34 degrees of freedom
Multiple R-squared:  0.9493,    Adjusted R-squared:  0.9478 
F-statistic:   637 on 1 and 34 DF,  p-value: < 2.2e-16

Challenge

Open exercises/ for prompts.

• Fit the Linear Model lm(hardness ~ dens, data = janka) to the janka data.

• Evaluate the model fit

Model diagnostics

plot(janka_ls, which=2)
plot(janka_ls, which=3)

Formal tests

Violations of assumptions

library(lmtest)

# Breusch-Pagan Test of Homoscedasticity
lmtest::bptest(janka_ls)

    studentized Breusch-Pagan test

data:  janka_ls
BP = 4.4668, df = 1, p-value = 0.03456

# Shapiro-Wilk test for normality of residuals
shapiro.test(residuals(janka_ls))

    Shapiro-Wilk normality test

data:  residuals(janka_ls)
W = 0.93641, p-value = 0.03933

Challenge

Open exercises/ for prompts.

• What can we try to improve the fit of our lm(hardness ~ dens, data = janka)

• Let’s make suggestions then try them out

BoxCox

Transformations

janka_sqrt <- lm(sqrt(hardness) ~ dens, data = janka)

plot(janka_sqrt, which=2)
plot(janka_sqrt, which=3)

Model fit

ggplot(janka, aes(x = hardness, y = dens)) +
  geom_point() +  # scatter plot of original data points
  geom_smooth(method = "lm", formula = y ~ sqrt(x)) +  # regression line
  labs(title = "Sqrt Linear Regression with ggplot2",
       x = "X", y = "Y")  # axis labels and title

Residual plots

janka_log <- lm(log(hardness) ~ dens, data = janka)

plot(janka_log, which=2)
plot(janka_log, which=3)

Polynomials

janka_poly <- lm(log(hardness) ~ poly(dens, 2), data = janka)

plot(janka_poly, which=2)
plot(janka_poly, which=3)

Summary polynomial

summary(janka_poly)

Call:
lm(formula = log(hardness) ~ poly(dens, 2), data = janka)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.22331 -0.05708 -0.01104  0.07500  0.18871 

Coefficients:
               Estimate Std. Error t value Pr(>|t|)    
(Intercept)      7.1362     0.0168 424.896  < 2e-16 ***
poly(dens, 2)1   3.3862     0.1008  33.603  < 2e-16 ***
poly(dens, 2)2  -0.5395     0.1008  -5.354 6.49e-06 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1008 on 33 degrees of freedom
Multiple R-squared:  0.9723,    Adjusted R-squared:  0.9706 
F-statistic: 578.9 on 2 and 33 DF,  p-value: < 2.2e-16

Troublesome transformations

ggplot(janka, aes(x = hardness, y = dens)) +
  geom_point() +  # scatter plot of original data points
  geom_smooth(method = "lm", formula = (y ~ log(x))) +  # regression line
  labs(title = "Log Linear Regression",
       x = "X", y = "Y")  # axis labels and title
ggplot(janka, aes(x = hardness, y = dens)) +
  geom_point() +  # scatter plot of original data points
  geom_smooth(method = "lm", formula = (y ~ poly(log(x), 2))) +  # regression line
  labs(title = "Quadratic Log Linear Regression",
       x = "X", y = "Y")  # axis labels and title

Weighted least squares

weights = 1/sqrt(hardness): This argument specifies the weights to be used in the fitting process. In this case, the weights are set to be inversely proportional to the square root of hardness.

1/sqrt(hardness): This means that observations with higher values of hardness will have lower weights, and observations with lower values of hardness will have higher weights. This could be used to give more importance to certain observations in the model fitting process. data = janka: This specifies the data frame (janka) from which the variables (sqrt(hardness) and dens) are to be taken.

Putting it all together, the code janka_wls <- lm(sqrt(hardness) ~ dens, weights = 1/sqrt(hardness), data = janka) fits a linear model where the response variable is the square root of hardness, the predictor variable is dens, and weights are inversely proportional to the square root of hardness. This is often referred to as a weighted least squares (WLS) regression model, where observations are weighted differently based on certain criteria (in this case, the square root of hardness).

janka_wls <- lm(sqrt(hardness) ~ dens, weights = 1/sqrt(hardness), data = janka)

plot(janka_wls, which=2)
plot(janka_wls, which=3)

Troublesome transformations

See screenshots

Plot model

ggplot(janka, aes(x = dens, y = hardness)) +
  geom_point() +  # scatter plot of original data points
  geom_smooth(method = "glm", method.args = list(gaussian(link = "sqrt"))) +  # regression line
  labs(title = "Linear Regression with ggplot2",
       x = "X", y = "Y")  # axis labels and title

Residuals

janka_glm <- glm(hardness ~ dens, data = janka, family = gaussian(link = "sqrt"))

plot(janka_glm, which=2)
plot(janka_glm, which=3)

Gamma distribution

Univariate analysis

summary statistics
------
min:  413   max:  3260 
median:  1195 
mean:  1469.472 
estimated sd:  801.5172 
estimated skewness:  0.6579162 
estimated kurtosis:  2.571426 

Gamma GLM

janka_gamma <- glm(hardness ~ dens, data = janka, family = Gamma(link = "sqrt"))

summary(janka_gamma)

Call:
glm(formula = hardness ~ dens, family = Gamma(link = "sqrt"), 
    data = janka)

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  1.86729    0.90357   2.067   0.0465 *  
dens         0.76780    0.02243  34.224   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for Gamma family taken to be 0.009716208)

    Null deviance: 11.26788  on 35  degrees of freedom
Residual deviance:  0.32876  on 34  degrees of freedom
AIC: 452.97

Number of Fisher Scoring iterations: 4

Plot model

ggplot(janka, aes(x = dens, y = hardness)) +
  geom_point() +  # scatter plot of original data points
  geom_smooth(method = "glm", method.args = list(Gamma(link = "sqrt"))) +  # regression line
  labs(title = "Linear Regression with ggplot2",
       x = "X", y = "Y")  # axis labels and title

Exercise

Challenge

Compare

library(MuMIn)

#r.squaredLR()

# r squared

Log-Likelihood

Maximum Likelihood

Deviance

Pseudo R-squared

AIC

Likelihood ratio Test

Challenge?

Binary variables

• Investigating mayfly abundance in relation to pollution gradient measured in Cumulative Criterion Units (CCU)

• Mayflies serve as indicators of stream health due to sensitivity to metal pollution

• Higher CCU values indicate increased metal pollution, potentially impacting mayfly presence or absence in the stream

Example

If I just fit a regular linear model to this data:

ggplot(Mayflies, aes(x=CCU, y=Occupancy)) + geom_point()+
  geom_smooth(method = "lm")
ggplot(Mayflies, aes(x=CCU, y=Occupancy)) + geom_point()+
  geom_smooth(method = "glm", method.args = list(binomial(link = "logit"))) +  # regression line
  labs(title = "Logit-link Binomial Regression",
       x = "X", y = "Y")  # axis labels and title

Bernoulli Distribution

Probability distribution

The link function

If \(μ = xβ\) is only true for normally distributed data, then, if this is not the case, we must use a transformation on the expected values:

\[g(μ) = xβ\]

where \(g(μ)\) is the link function.

This allows us to relax the normality assumption.

The logit link

For binary data, the link function is called the logit:

\[ logit(p) = \log \frac{p}{1 - p} \]

where \(p\) represents the expected values (probability that \(Y = 1\)).

To obtain the odds \((p / (1 - p))\), we log-transform them.

The logit link

\[ logit(p) = \log \frac{p}{1 - p} \]

Get the odds \((μ / (1 - μ))\) and log-transform them.

The odds put our expected values on a 0 to +Inf scale.

The log transformation puts our expected values on a -Inf to +Inf scale.

Now, the expected values can be linearly related to the linear predictor.

Probability, odds, logit-odds

Example

mayfly_glm <- glm(Occupancy ~ CCU, family = binomial(link = "logit"), data = Mayflies)

summary(mayfly_glm)

Call:
glm(formula = Occupancy ~ CCU, family = binomial(link = "logit"), 
    data = Mayflies)

Coefficients:
            Estimate Std. Error z value Pr(>|z|)  
(Intercept)    5.102      2.369   2.154   0.0313 *
CCU           -3.051      1.211  -2.520   0.0117 *
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 34.795  on 29  degrees of freedom
Residual deviance: 12.649  on 28  degrees of freedom
AIC: 16.649

Number of Fisher Scoring iterations: 7

Each coefficient corresponds to a predictor variable, indicating the change in the log odds of the response variable associated with a one-unit change in that predictor, holding other predictors constant.

Interpret the model

• Interpretation of coefficients involves exponentiating them to obtain odds ratios.

• An odds ratio greater than 1 implies higher odds of the event occurring with an increase in the predictor.

• An odds ratio less than 1 implies lower odds of the event occurring with an increase in the predictor.

Intepret the model

Logit odds

When we use a binomial model, we produce the ‘log-odds’, this produces a fully unconstrained linear regression as anything less than 1, can now occupy an infinite negative value -∞ to ∞.

\[\log\left(\frac{p}{1-p}\right) = \beta_{0}+\beta_{1}x_{1}+\beta_{2}x_{2}\]

\[\frac{p}{1-p} = e^{\beta_{0}}e^{\beta_{1}x_{1}}e^{\beta_{2}x_{2}}\]

• We can interpret \(\beta_{1}\) and \(\beta_{2}\) as the increase in the log odds for every unit increase in \(x_{1}\) and \(x_{2}\).

• We could alternatively interpret \(\beta_{1}\) and \(\beta_{2}\) using the notion that a one unit change in \(x_{1}\) as a percent change of \(e^{\beta_{1}}\) in the odds.

For the intercept: - The estimate as log odds is 5.102. - Therefore the odds are 164 - Over 99% probability of observing mayfly when CCU = 0

Challenge

Open exercises/ for prompts.

• Work with the odds and probability calculators

• Check how comfortable you are with additive log-odds, multiplicative odds and changing probability

Model fit

For a simple Bernoulli/Binary GLM there are only a few checks that apply:

Model fit

Model fit

DHARMa fit

Prediction

           1            2            3            4            5            6 
9.939504e-01 8.860420e-01 2.689747e-01 1.711402e-02 8.233054e-04 3.899164e-05 

Emmeans

 CCU     prob       SE  df asymp.LCL asymp.UCL
   0 0.993950 0.014243 Inf  6.13e-01    0.9999
   1 0.886042 0.128537 Inf  3.91e-01    0.9895
   2 0.268975 0.147493 Inf  7.80e-02    0.6155
   3 0.017114 0.026259 Inf  8.16e-04    0.2707
   4 0.000823 0.002214 Inf  4.20e-06    0.1387
   5 0.000039 0.000151 Inf  0.00e+00    0.0714

Confidence level used: 0.95 
Intervals are back-transformed from the logit scale 

Prediction plots with emmeans

Exercise

Challenge

Open exercises/ for prompts.

• Using the malaria dataset fit a binomial glm to look at the predictors of malaria in the Seychelles warbler

• Check the model fit

• Make predictions

• Produce a suitable figure

Proportion data and GLM

Sometimes, count data aligns more closely with logistic regression methodologies than it initially appears.

We’re not looking at typical count data when measuring the number of occurrences with a known total sample size.

Imagine, for example, we’re studying the prevalence of a native underbrush species across various forest plots. We assess 15 different plots, each meticulously surveyed for the presence of this underbrush, counting the number of square meters where the underbrush is present versus absent within a standard area:

\[\frac{M^2~\text{with native underbrush (successes)}}{\text{Total surveyed area in}~M^2~(\text{trials})}\]

Bound between zero and one

Binomial distribution

\(P(X = s) = C(n, s) \cdot p^s \cdot (1 - p)^{n - s}\)

Where:

• \(n\) is the total number of trials of an event.
• \(s\) corresponds to the number of times an event should occur.
• \(p\) is the probability that the event will occur.
• \((1 - p)\) is the probability that the event will not occur.
• \(C\) term represents combinations, calculated as \(C(n, s) = \frac{n!}{s!(n - s)!}\), representing the number of ways to choose \(s\) successes out of \(n\) trials.

Example

TEXT

:::: columns

# According to the situation:
# n = 5,
# s = 5,
# p = 0.95,
# (1 — p) = 0.05

p = 1 * (0.95)^(5) * (0.05)^(0)

p

[1] 0.7737809

[1] 0.7737809

Case study

Binomial VS Bernoulli Keypoints!

Bernoulli deals with the outcome of the single trial of the event, whereas Binomial deals with the outcome of the multiple trials of the single event.

Bernoulli is used when the outcome of an event is required for only one time, whereas the Binomial is used when the outcome of an event is required multiple times.

Exercise

Weights

Predictions/write-up

Overdispersion

Where:

• \(n\) is the total number of trials of an event.

• \(s\) corresponds to the number of times an event should occur.

• \(p\) is the probability that the event will occur.

• \((1 - p)\) is the probability that the event will not occur.

• \(C\) term represents combinations, calculated as \(C(n, s) = \frac{n!}{s!(n - s)!}\), representing the number of ways to choose \(s\) successes out of \(n\) trials.

• \(\theta\) term describes additional variance outside of the Binomial distribution

Challenge

Poisson

GLM

GLM Poisson

Poisson limitations

The Poisson distribution has the following characteristics:

Poisson: Case study

This model is inadequate. It is predicting negative begging calls within the range of the observed data, which clearly does not make any sense.

cuckoo_lm <- lm(Beg ~ Mass + Species + Mass:Species, data = cuckoo)

Diagnostics

Let us display the model diagnostics plots for this linear model.

• Curvature

• Funnelling effect

Biological data and distributions

fitdistrplus?

Poisson model

We should therefore try a different model structure.

The response variable in this case is a classic count data: discrete and bounded below by zero (i.e we cannot have negative counts). We will therefore try a Poisson model using the canonical log link function for the mean:

• λ varies with x (mass) which means residual variance will also vary with x, which means that we just relaxed the homogeneity of variance assumption!

• Predicted values will now be integers instead of fractions

• The model will never predict negative values (Poisson is strictly positive)

\[ \log{\lambda} = \beta_0 + \beta_1 M_i + \beta_2 S_i + \beta_3 M_i S_i \]

where \(M_i\) is nestling mass and \(S_i\) a dummy variable

\(S_i = \left\{\begin{array}{ll}1 & \text{if } i \text{ is warbler},\\0 & \text{otherwise}.\end{array}\right.\)

The term \(M_iS_i\) is an interaction term. Think of this as an additional explanatory variable in our model. Effectively it lets us have different slopes for different species (without an interaction term we assume that both species have the same slope for the relationship between begging rate and mass, and only the intercept differ).

Regression lines

The mean regression lines for the two species look like this:

Fit the Poisson model

Fit the model with the interaction term in R:

cuckoo_glm1 <- glm(Beg ~ Mass + Species + Mass:Species, data=cuckoo, family=poisson(link="log"))

summary(cuckoo_glm1)

Call:
glm(formula = Beg ~ Mass + Species + Mass:Species, family = poisson(link = "log"), 
    data = cuckoo)

Coefficients:
                     Estimate Std. Error z value Pr(>|z|)    
(Intercept)          0.334475   0.227143   1.473  0.14088    
Mass                 0.094847   0.007261  13.062  < 2e-16 ***
SpeciesWarbler       0.674820   0.259217   2.603  0.00923 ** 
Mass:SpeciesWarbler -0.021673   0.008389  -2.584  0.00978 ** 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for poisson family taken to be 1)

    Null deviance: 970.08  on 50  degrees of freedom
Residual deviance: 436.05  on 47  degrees of freedom
AIC: 615.83

Number of Fisher Scoring iterations: 6

Note there appears to be a negative interaction effect for Species:Mass, indicating that Begging calls do not increase with mass as much as you would expect for Warblers as compared to Cuckoos.

Exercise

Open exercises/exercise_04.R for prompts.

• Fit a Poisson model to the cuckoo data

• Look at the residual plots - how have they changed?

Model summary

Parameter estimates

Deviance

Poisson model check

Mean-centering

Model selection

Likelihood ratio tests, AIC - exercise/reminder?

Exercise

Look at the effect of mean-centering on estimates

Offset - for rates

Call:
glm(formula = Call_Total ~ mass.c + Species + mass.c:Species, 
    family = poisson(link = "log"), data = cuckoo, offset = log(Mins))

Coefficients:
                       Estimate Std. Error z value Pr(>|z|)    
(Intercept)            2.249441   0.030831  72.961   <2e-16 ***
mass.c                 0.129261   0.002502  51.671   <2e-16 ***
SpeciesWarbler         0.300202   0.035330   8.497   <2e-16 ***
mass.c:SpeciesWarbler -0.057893   0.002829 -20.462   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for poisson family taken to be 1)

    Null deviance: 9597.9  on 50  degrees of freedom
Residual deviance: 3071.8  on 47  degrees of freedom
AIC: 3334.6

Number of Fisher Scoring iterations: 6

Poisson limitations

So for the Poisson regression case we assume that the mean and variance are the same. Hence, as the mean increases, the variance increases also (heteroscedascity). This may or may not be a sensible assumption so watch out! Just because a Poisson distribution usually fits well for count data, doesn’t mean that a Gaussian distribution can’t always work.

Recall the link function between the predictors and the mean and the rules of logarithms (if \(\log{\lambda} = k\)(value of predictors), then \(\lambda = e^k\)):

Overdispersion

When the residual deviance is higher than the residual degrees of freedom, we say that the model is overdispersed. This situation is mathematically represented as:

\[ \phi = \frac{\text{Residual deviance}}{\text{Residual degrees of freedom}} \]

Overdispersion occurs when the variance in the data is even higher than the mean, indicating that the Poisson distribution might not be the best choice. This can be due to many reasons, such as the presence of many zeros, many very high values, or missing covariates in the model.

Solutions

Quasi-Poisson GLM

The systematic part and the link function remain the same

\[ \log{\lambda} = \beta_0 + \beta_1 x_i \]

phi\(\phi\) is a dispersion parameter, it will not affect estimates but will affect significance. Standard errors are multiplied by \(\sqrt\phi\)

\[ Y_i = Poisson(\phi \lambda_i) \]

Where:

• \(Yi\) is the response variable.
• \(\phi\) is the dispersion parameter, which adjusts the variance of the distribution

Confidence intervals and p-values WILL change.

Exercise

Open exercises/exercise_04.R for prompts.

• Fit a quasipoisson model to the cuckoo data or update the previous one

• Look at the residual plots - how have they changed?

• Look at the SE and p-values - how have they changed

• Calculate new 95% confidence intervals

Summary

Deviance analysis

F distributions

The presence of overdispersion suggested the use of the F-test for nested models. We will test if the squared term can be dropped from the model.

https://towardsdatascience.com/the-most-important-statistical-test-dee01f4d50cf

Zero-inflation

“Zero-inflation” refers to the problem of “too many zeros”.

The dependent variable contains more zeros than would be expected under the standard statistical distribution for zero-bounded count data (Poisson or negative binomial).

Zero-inflation is a common feature of count datasets. Large numbers of zeros can arise because

1. There was nothing there to count (true zeros)

2. Things were there but missed (false zeros)

Zero-inflation

ZIP in R

In a ZIP model we define two halves of the model; the first half models the count process (i.e. a Poisson GLM explaining non-zero integers) and the second half models the binomial (i.e. explains the zeros).

install.packages("pscl")

zip1 <- zeroinf(y ~ x1 + x2 | x1 + x2,
                dist = "poisson",
                link = "logit",
                data = dataframe)

Alternative model design

You can change the zero-inflation model’s link function and use AIC to decide on the best model

# zeros modelled with a constant
y ~ x1 + x2

# zeros modelled with the same variables
y ~ x1 + x2 | x1 + x2

# zeros modelled with different variables
y ~ x1 + x2 | z1 +z2

Checking for zero-inflation

RETURN TO CUCKOO DATA!

Fit the model

AIC

Diagnostics

qqnorm(sresid) qqline(sresid, col = “red”)

Negative Binomial

Negative Binomial

Negative binomial GLMs are favored when overdispersion is high.

It has two parameters, \(μ\) and \(k\). \(k\) controls for the dispersion parameter (smaller \(k\) indicates higher dispersion). It corresponds to a combination of two distributions (Poisson and gamma).

It assumes that the \(Y_i\) are Poisson distributed with the mean \(μ_i\) assumed to follow a gamma distribution:

\[ E(Y_i) = μ_i \\ \text{Var}(Y_i) = μ_i + μ_i^2/k \]

Fitting a negative binomial in R

Negative binomial is not in the glm() function

Exercise

Challenge

Gamma

Where next?

Live demo!

Exercise 4

Open exercises/exercise_04.R for prompts.

• Look at the ACF and PACF plots of the residuals from your previous GAM.

• Try fitting a GAMM model instead.

Additional resources

• GLMs resource list:

• Discovering Statistics - Andy Field

• An Introduction to Generalized Linear Models - Dobson & Barnett

• An Introduction to Statistical Learning with Applications in R - James, Witten, Hastie & Tibshirani

• Mixed Effects Models and Extensions in Ecology with R - Zuur, et al.